the circle, then of the circumferences which they intercept, that which is the nearer to the given point is less than the other.

FDCG, let there be drawn BFG, BDC, cutting the circumference in the points F, G, and D, C, respectively: Then is FD <GC.

First, let one of the straight lines drawn from B, as BC, pass through the centre K of the circle: Join K, F and K, G; then (E. 16. 1.) the exterior GKC, of the A GKB, is > the but (E. 15. def. 1. and E. 5. 1.) the 4 KFG;.. the GKC > the 16. 1.) the KFG> the FKB or FKD; much more, then, is the GKC> <FKD; .. (S. 37. 3.)

CG > FD; i. e. FD < GC.

KGF;
KGF=

KFG; and (E.

But if BLM do not pass through the centre, find (E. 1. 3.) the centre K; join B, K; and produce it to meet the circumference in C: Then it

may be shewn, as before, that FD < GC, and that

DL < CM; .. FDL < GCM.

PROP. XXXIX.

51. THEOREM. The straight lines joining the extremities of the chords of two equal arches of the same circle, toward the same parts, are parallel to each other.

Let AC, BG be the chords of two equal arches

B

G

AC, BG, of the circle ABGC; and let A, B, and C, G be joined: Then CG is parallel to AB.

For join C, B; and since (hyp.) AC=BG,.. (E. 27. 3.) the ABC= 4 BCG; .. (E. 27. 1.) CG is parallel to AB.

PROP. XL.

52. THEOREM. In equal circles, the greater of two circumferences subtends the greater angle, whether the angles compared be at the centres or the circumferences.

For if not, the standing on the greater circumference is equal to the other or less than it; but it cannot be equal; for then (E. 26. 3.) the two circumferences would be equal, which is contrary to the hypothesis: Neither can it be less, for, then, (S. 37. 3.) the greater circumference would be less than the other, which is absurd. Wherefore, the greater of two circumferences subtends the greater 4, whether the two be at the centres or circumferences.

PROP. XLI.

53. PROBLEM. If any equilateral rectilineal figure, of an even number of sides, be inscribed in a given circle, to find a curvilineal figure that is equal to it, and that is bounded by arches of circles, each of which circles is equal to the given circle.

Let ABCDEF be an equilateral rectilineal figure, of an even number of sides, inscribed in the given circle ACE; It is required to find a

curvilineal figure equal to it, and bounded by arches of circles that are equal to the given circle ACE.

On half the number of sides of the inscribed figure, taken contiguous to one another, as BC, CD, DE, describe (S. 38. 3.) segments of circles, BGC, CHD, DIE, each similar to the segment cut off from the given circle by each of the sides: The curvilineal figure contained by BGC, CHD, DIE, EF, and FA, and AB, is equal to the inscribed polygon ABCDEF.

For, since (hyp.) the figure is equilateral, (E. 28. 3.) AB=BC; .. AFEDCB = BAFEDC; . (E. 27. 3.) the in the segment cut off by AB is equal to the in the segment cut off by BC; ..these two segments (E. 11. def. 3.) are similar, and (hyp. and E. 24.3.) equal to one another.

And, in the same manner, may all the segments, cut off by the equal sides of the inscribed figure, be shewn to be similar and equal to one another, and to the segments BGC, CHD, DIE.

But the figure contained by BA, AF, FE, EID, DHC and CGB, together with the segments BGC, CHD, DIE, makes up the equilateral rectilineal figure ABCDEF; and that same figure, together with the equal segments cut off by BA, AF, and FE, makes up the curvilineal figure contained by BGC, CHD, DIE, EF, FA and AB; the which figure is, ..., equal to the inscribed rectilineal figure ABCDEF.*

PROP. XLII.

54. THEOREM. In equal circles, the greater chord subtends the greater circumference.

For (hyp. and E. 15. def. 1. and E. 25. 1.) the 4 subtended, at the centre, by the greater chord is the subtended, at the centre, by the less: .. (S. 37. 3.) the circumference subtended by the

* It is easy to shew, by the help, chiefly, of S. 7. 3., that when the equilateral figure inscribed in the circle, is a square, the circumferences of the similar segments, described in the course of the demonstration, touch one another in the common. extremity of the two contiguous sides; and that when the inscribed polygon has any greater number of sides, as six, eight, &c., the circumferences of any two of the segments meeting one another in the common extremity of two contiguous sides, do not meet again within the circle.